EECS 142
Lecture 8: Distortion Metrics
Prof. Ali M. Niknejad
University of California, Berkeley
c 2005 by Ali M. Niknejad
Copyright A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 1/26
– p. 1
Output Waveform
In general, then, the output waveform is a Fourier series
vo = V̂o1 cos ω1 t + V̂o2 cos 2ω1 t + V̂o3 cos 3ω1 t + . . .
V̂o
100mV
10mV
1mV
Gain Compression
Higher Order
Distortion Products
100µV
V̂o2
10µV
V̂o3
1µV
1µV
A. M. Niknejad
10µV
100µV
1mV
University of California, Berkeley
10mV
V̂i
EECS 142 Lecture 8 p. 2/26
– p. 2
Fractional Harmonic Distortion
The fractional second-harmonic distortion is a
commonly cited metric
ampl of second harmonic
HD2 =
ampl of fund
If we assume that the square power dominates the
second-harmonic
HD2 =
or
HD2 =
A. M. Niknejad
S12
a2 2
a1 S1
1 a2
2 a S1
1
University of California, Berkeley
EECS 142 Lecture 8 p. 3/26
– p. 3
Third Harmonic Distortion
The fractional third harmonic distortion is given by
ampl of third harmonic
HD3 =
ampl of fund
If we assume that the cubic power dominates the third
harmonic
S12
a3 4
HD3 =
a1 S1
or
1 a3 2
HD3 =
S1
4 a1
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 4/26
– p. 4
Output Referred Harmonic Distortion
In terms of the output signal Som , if we again neglect
gain expansion/compression, we have Som = a1 S1
1 a2
HD2 =
Som
2
2 a1
1 a3 2
HD3 =
Som
3
4 a1
On a dB scale, the second harmonic increases linearly
with a slope of one in terms of the output power
whereas the thrid harmonic increases with a slope of 2.
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 5/26
– p. 5
Signal Power
Recall that a general memoryless non-linear system will
produce an output that can be written in the following
form
vo (t) = V̂o1 cos ω1 t + V̂o2 cos 2ω1 t + V̂o3 cos 3ω1 t + . . .
By Parseval’s theorem, we know the total power in the
signal is related to the power in the harmonics
Z
Z X
X
V̂oj cos(jω1 t)
v 2 (t)dt =
V̂ok cos(kω1 t)dt
T
T
=
XXZ
j
A. M. Niknejad
j
k
T
k
V̂oj cos(jω1 t)V̂ok cos(kω1 t)dt
University of California, Berkeley
EECS 142 Lecture 8 p. 6/26
– p. 6
Power in Distortion
By the orthogonality of the harmonics, we obtain
Parseval’s Them
Z
XX
X
2
1
1
V̂
V̂
=
v 2 (t)dt =
δ
V̂
oj
ok
jk
oj
2
2
T
j
k
j
The power in the distortion relative to the fundamental
power is therefore given by
2
2
Vo3
Power in Distortion
Vo2
= 2 + 2 + ···
Power in Fundamental Vo1 Vo1
= HD22 + HD32 + HD42 + · · ·
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 7/26
– p. 7
Total Harmonic Distortion
We define the Total Harmonic Distortion (T HD) by the
following expression
q
T HD = HD22 + HD32 + · · ·
Based on the particular application, we specify the
maximum tolerable T HD
Telephone audio can be pretty distorted (T HD < 10%)
High quality audio is very sensitive (T HD < 1% to
T HD < .001%)
Video is also pretty forgiving, T HD < 5% for most
applications
Analog Repeaters < .001%. RF Amplifiers < 0.1%
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 8/26
– p. 8
Intermodulation Distortion
So far we have characterized a non-linear system for a
single tone. What if we apply two tones
Si = S1 cos ω1 t + S2 cos ω2 t
So = a1 Si + a2 Si2 + a3 Si3 + · · ·
= a1 S1 cos ω1 t + a1 S2 cos ω2 t + a3 (Si )3 + · · ·
The second power term gives
a2 S12 cos2 ω1 t + a2 S22 cos2 ω2 t + 2a2 S1 S2 cos ω1 t cos ω2 t
S12
S22
= a2 (cos 2ω1 t + 1) + a2 (cos 2ω2 t + 1) +
2
2
a2 S1 S2 (cos(ω1 + ω2 )t − cos(ω1 − ω2 )t)
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 9/26
– p. 9
Second Order Intermodulation
The last term cos(ω1 ± ω2 )t is the second-order
intermodulation term
The intermodulation distortion IM2 is defined when the
two input signals have equal amplitude Si = S1 = S2
Amp of Intermod a2
IM2 =
= Si
Amp of Fund
a1
Note the relation between IM2 and HD2
IM2 = 2HD2 = HD2 + 6dB
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 10/26
– p.
Practical Effects of IM2
This term produces distortion at a lower frequency
ω1 − ω2 and at a higher frequency ω1 + ω2
Example: Say the receiver bandwidth is from
800MHz − 2.4GHz and two unwanted interfering signals
appear at 800MHz and 900MHz.
Then we see that the second-order distortion will
produce distortion at 100MHz and 1.7GHz. Since 1.7GHz
is in the receiver band, signals at this frequency will be
corrupted by the distortion.
A weak signal in this band can be “swamped” by the
distortion.
Apparently, a “narrowband” system does not suffer from
IM2 ? Or does it ?
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 11/26
– p.
Low-IF Receiver
In a low-IF or direct conversion receiver, the signal is
down-converted to a low intermediate frequency fIF
Since ω1 − ω2 can potentially produce distortion at low
frequency, IM2 is very important in such systems
Example: A narrowband system has a receiver
bandwidth of 1.9GHz - 2.0GHz. A sharp input filter
eliminates any interference outside of this band. The IF
frequency is 1MHz
Imagine two interfering signals appear at f1 = 1.910GHz
and f2 = 1.911GHz. Notice that f2 − f1 = fIF
Thus the output of the amplifier/mixer generate
distortion at the IF frequency, potentially disrupting the
communication.
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 12/26
– p.
Cubic IM
Now let’s consider the output of the cubic term
a3 s3i = a3 (S1 cos ω1 t + S2 cos ω2 t)3
Let’s first notice that the first and last term in the
expansion are the same as the cubic distortion with a
single input
3
a3 S1,2
4
(cos 3ω1,2 t + 3 cos ω1,2 t)
The cross terms look like
3
a3 S1 S22 cos ω1 t cos2 ω2 t
2
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 13/26
– p.
Third Order IM
Which can be simplified to
3
3 cos ω1 t cos ω2 t = cos ω1 t(1 + cos 2ω2 t) =
2
2
3
3
= cos ω1 t + cos(2ω2 ± ω1 )
2
4
The interesting term is the intermodulation at 2ω2 ± ω1
By symmetry, then, we also generate a term like
3
2
a3 S1 S2
4
cos(2ω1 ± ω2 )
Notice that if ω1 ≈ ω2 , then the intermodulation
2ω2 − ω1 ≈ ω1
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 14/26
– p.
Inband IM3 Distortion
S(ω)
Interfering Signals
wanted
ω3
2ω1
distortion product
ω
ω1 ω2
ω2
2ω2
ω1
Now we see that even if the system is narrowband, the
output of an amplifier can contain in band
intermodulation due to IM3 .
This is in contrast to IM2 where the frequency of the
intermodulation was at a lower and higher frequency.
The IM3 frequency can fall in-band for two in-band
interferer
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 15/26
– p.
Definition of IM3
We define IM3 in a similar manner for Si = S1 = S2
Amp of Third Intermod 3 a3 2
=
IM3 =
Si
Amp of Fund
4 a1
Note the relation between IM3 and HD3
IM3 = 3HD3 = HD3 + 10dB
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 16/26
– p.
Complete Two-Tone Response
S(ω)
ω1
−
−− −−
−
−−
−
ω2
ω2
ω1
2ω2 2ω1
ω1 + ω2
ω1 ω2
3ω1
2ω1
2ω2
ω2
3ω1
2ω2
ω1
3ω2
2ω1 2ω2
ω2 3ω2
ω1
ω
3ω1 3ω2
2ω1 + ω2
2ω2 + ω1
2ω1
We have so far identified the harmonics and IM2 and
IM3 products
A more detailed analysis shows that an order n
non-linearity can produce intermodulation at
frequencies jω1 ± kω2 where j + k = n
All tones are spaced by the difference ω2 − ω1
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 17/26
– p.
Distortion of AM Signals
Consider a simple AM signal (modulated by a single
tone)
s(t) = S2 (1 + m cos ωm t) cos ω2 t
where the modulation index m ≤ 1. This can be written
as
m
m
s(t) = S2 cos ω2 t + cos(ω2 − ωm )t + cos(ω2 + ωm )t
2
2
The first term is the RF carrier and the last terms are
the modulation sidebands
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 18/26
– p.
Cross Modulation
Cross modulation occurs in AM systems (e.g. video
cable tuners)
The modulation of a large AM signal transfers to
another carrier going thru the same amp
Si = S1 cos ω1 t + S2 (1 + m cos ωm t) cos ω2 t
| {z } |
{z
}
wanted
interferer
CM occurs when the output contains a term like
K(1 + δ cos ωm t) cos ω1 t
Where δ is called the transferred modulation index
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 19/26
– p.
Cross Modulation (cont)
For So = a1 Si + a2 Si2 + a3 Si3 + · · · , the term a2 Si2 does
not produce any CM
The term
a3 Si3 = · · · + 3a3 S1 cos ω1 t (S2 (1 + m cos ωm t) cos ω2 t)2 is
expanded to
= · · · + 3a3 S1 S22 cos ω1 t(1 + 2m cos ωm t + m2 cos2 ωm t)×
1
2 (1 + cos 2ω2 t)
Grouping terms we have in the output
a3 2
So = · · · + a1 S1 (1 + 3 S2 m cos ωm t) cos ω1 t
a1
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 20/26
– p.
CM Definition
unmodulated waveform (input)
modulated waveform due to CM
Transferred Modulation Index
CM =
Incoming Modulation Index
a3 2
CM = 3 S2 = 4IM3
a1
= IM3 (dB) + 12dB
= 12HD3 = HD3 (dB) + 22dB
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 21/26
– p.
Distortion of BJT Amplifiers
VCC
Consider the CE BJT
amplifier shown. The
biasing is omitted for
clarity.
RL
vo
+
vs
The output voltage is simply
Vo = VCC − IC RC
Therefore the distortion is generated by IC alone.
Recall that
IC = IS eqVBE /kT
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 22/26
– p.
BJT CE Distortion (cont)
Now assume the input VBE = vi + VQ , where VQ is the
bias point. The current is therefore given by
VQ
VT
IC = IS e e
| {z }
vi
VT
IQ
Using a Taylor expansion for the exponential
1 2 1 3
e = 1 + x + x + x + ···
2!
3!
2
3
1 vi
1 vi
vi
+
+
+ ···)
IC = IQ (1 +
VT
2 VT
6 VT
x
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 23/26
– p.
BJT CE Distortion (cont)
Define the output signal ic = IC − IQ
q 3
IQ
1 q 2
1
vi +
IQ vi2 +
IQ vi3 + · · ·
ic =
VT
2 kT
6 kT
Compare to So = a1 Si + a2 Si2 + a3 Si3 + · · ·
qIQ
a1 =
= gm
kT
1 q 2
IQ
a2 =
2 kT
1 q 3
a3 =
IQ
6 kT
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 24/26
– p.
Example: BJT HD2
For any BJT (Si, SiGe, Ge, GaAs), we have the
following result
1 q vˆi
HD2 =
4 kT
where vˆi is the peak value of the input sine voltage
For vˆi = 10mV, HD2 = 0.1 = 10%
We can also express the distortion as a function of the
output current swing iˆc
1 iˆc
1 a2
Som =
HD2 =
2
2 a1
4 IQ
For
A. M. Niknejad
iˆc
IQ
= 0.4, HD2 = 10%
University of California, Berkeley
EECS 142 Lecture 8 p. 25/26
– p.
Example: BJT IM3
Let’s see the maximum allowed signal for IM3 ≤ 1%
3 a3 2 1
S1 =
IM3 =
4 a1
8
q vˆi
kT
2
Solve vˆi = 7.3mV. That’s a pretty small voltage. For
practical applications we’d like to improve the linearity of
this amplifier.
A. M. Niknejad
University of California, Berkeley
EECS 142 Lecture 8 p. 26/26
– p.